Matematika/Gauso formulė: Skirtumas tarp puslapio versijų
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39 eilutė:
:<math>=\iint_S x^3 \sqrt{1+ \frac{y^2}{R^2-y^2-z^2}+ \frac{ z^2}{R^2-y^2-z^2} } \mathbf{d}y \mathbf{d}z=</math>
:<math>=\iint_S x^3 \sqrt{ \frac{R^2-y^2-z^2 +y^2+z^2}{R^2-y^2-z^2} } \mathbf{d}y \mathbf{d}z=</math>
:<math>=\iint_S x^3 \sqrt{ \frac{R^2}{R^2-y^2-z^2} } \mathbf{d}y \mathbf{d}z=\iint_S \sqrt{R^2-y^2-z^2} \sqrt{ \frac{R^2}{R^2-y^2-z^2} } \mathbf{d}y \mathbf{d}z=\iint_S R \mathbf{d}y \mathbf{d}z=</math>
:<math>=\iint_S
:<math>=\int_0^{2\pi} \left(
:Pasinaudodami [http://integrals.wolfram.com/index.jsp?expr=%28R%5E2+-+x%5E2%29%5E%283%2F2%29&random=false internetiniu integratoriumi], gauname, kad
:<math> \int_0^R \sqrt{(R^2-\rho^2)^3} \mathbf{d}\rho =\frac{1}{8}\left( \rho(5 R^2 - 2\rho^2)\sqrt{R^2-\rho^2}+3R^4 \arctan\frac{\rho}{\sqrt{R^2-\rho^2}} \right)|_0^R.</math>
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